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Thread: challenge question -- Factor the polynomial completely

  1. #1
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    challenge question -- Factor the polynomial completely

    Edit:

    Demonstrate at least two methods for factoring the following polynomial
    completely over the integers.


    [tex]x^5 + x^4 + x^3 + x^2 + x + 1[/tex]
    Last edited by lookagain; 01-25-2013 at 05:12 PM. Reason: My original post was awkward.

  2. #2
    Full Member MarkFL's Avatar
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    Method 1:

    Factor first by grouping:

    [tex](x^5+x^4)+(x^3+x^2)+(x+1)=x^4(x+1)+x^2(x+1)+(x+1)= (x+1)(x^4+x^2+1)[/tex]

    Now, for the quartic factor assume it may be factored as follows:

    [tex]x^4+x^2+1=(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2) x^2+(a+b)x+1[/tex]

    Equating coefficients, we find:

    [tex]a+b=0[/tex]

    [tex]ab+2=1[/tex]

    and so one solution is [tex](a,b)=(1,-1)[/tex] and we have:

    [tex]x^4+x^2+1=(x^2+x+1)(x^2-x+1)[/tex] which means:

    [tex]x^5+x^4+x^3+x^2+x+1=(x+1)(x^2+x+1)(x^2-x+1)[/tex]

    Method 2:

    Let:

    [tex]S=x^5+x^4+x^3+x^2+x+1[/tex] and so:

    [tex]Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1[/tex] hence:

    [tex]S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)[/tex] thus:

    [tex]S=(x+1)(x^2-x+1)(x^2+x+1)[/tex]
    Living in the pools, They soon forget about the sea...— Rush, "Natural Science" (1980)

  3. #3
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    Quote Originally Posted by lookagain View Post
    A) Give the completely factored form over the integers of the following polynomial, and

    B) demonstrate at least two methods for doing so.


    [tex]x^5 + x^4 + x^3 + x^2 + x + 1[/tex]


    (x+1)(x4+x2+1) → (x+1)(x4+2x2+1 - x2) → (x+1)(x2+1+x)(x2+1-x)

    or

    (x3+1)(x2+x+1) →(x+1)(x2-x+1)(x2+x+1)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  4. #4
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    Here is another:


    (x^5 + 1) + (x^4 + x^3 + x^2 + x) =

    (x + 1)(x^4 - x^3 + x^2 - x + 1) + (x^4 + x^3) + (x^2 + x) =

    (x + 1)(x^4 - x^3 + x^2 - x + 1) + x^3(x + 1) + x(x + 1) =

    (x + 1)(x^4 - x^3 + x^2 - x + 1 + x^3 + x) =

    (x + 1)(x^4 + x^2 + 1) =



    (You can continue simplifying this with one of the above steps in
    any of the appropriate above posts.)
    Last edited by lookagain; 01-25-2013 at 05:33 PM.

  5. #5
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    Here's another way.

    Since [tex]x^5+x^4+x^3+x^2+x+1 = \dfrac{x^6-1}{x-1}[/tex] the roots of this polynomial are exactly the set [tex]\{z\in \mathbb{C}-\{1\}\,\,;\,\, z^6=1\}[/tex], i.e. the roots of unity, ignoring the positive real root. They are [tex]e^{\pm i\pi/3},e^{\pm i2\pi/3}, -1 [/tex].

    We want a real factorization obviously, and we can see that the conjugate pair to each root is present (as it should be). Pairing them off we get the (minimal) polynomials for each:

    [tex](x-e^{i\pi/3})(x-e^{- i\pi/3}) = x^2-x+1[/tex]
    [tex](x-e^{i2\pi/3})(x-e^{- i2\pi/3}) = x^2+x+1[/tex]
    [tex]x-(-1) = x+1[/tex]

    There is an abstract algebra/number theoretic variation of the above that can be performed for the general case too.

  6. #6
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    Hello, lookagain

    This is a variation of daon's solution.


    Demonstrate at least two methods for factoring the following polynomial:

    . . . [tex]P(x) \;=\;x^5 + x^4 + x^3 + x^2 + x + 1[/tex]

    [tex]P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1} [/tex]

    . . . . .[tex]=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}[/tex]

    . . . . .[tex]=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}[/tex]

    . . . . .[tex]=\; (x+1)(x^2+x+1)(x^2-x+1)[/tex]

  7. #7
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    Quote Originally Posted by soroban View Post
    Hello, lookagain

    This is a variation of daon's solution.



    [tex]P(x) \;=\;\dfrac{x^6 - 1}{x-1} \;=\;\dfrac{\overbrace{(x^3)^2 - (1^2)}^{\text{diff. of squares}}}{x-1} [/tex]

    . . . . .[tex]=\;\dfrac{\overbrace{(x^3-1)}^{\text{diff.of cubes}}\cdot\overbrace{(x^3+1)}^{\text{sum of cubes}}}{x-1}[/tex]

    . . . . .[tex]=\; \dfrac{(\color{red}{\rlap{/////}}{x-1})(x^2+x+1)(x+1)(x^2-x+1)}{\color{red}{\rlap{/////}}x-1}[/tex]

    . . . . .[tex]=\; (x+1)(x^2+x+1)(x^2-x+1)[/tex]
    Quote Originally Posted by MarkFL

    Method 2:

    Let:

    [tex]S=x^5+x^4+x^3+x^2+x+1[/tex] and so:

    [tex]Sx=x^6+x^5+x^4+x^3+x^2+x=S+x^6-1[/tex] hence:

    [tex]S(x-1)=x^6-1=(x^3+1)(x^3-1)=(x+1)(x^2-x+1)(x-1)(x^2+x+1)[/tex] thus:

    [tex]S=(x+1)(x^2-x+1)(x^2+x+1)[/tex]

    These two (MarkFL's and soroban's versions) look essentially the same to me.


    - - - - - - - - - - - - - - - - -


    Others:


    (x^5 + x^2) + (x^4 + x) + (x^3 + 1) =

    x^2(x^3 + 1) + x(x^3 + 1) + 1(x^3 + 1) =

    (x^3 + 1)(x^2 + x + 1) =

    (x + 1)(x^2 - x + 1)(x^2 + x + 1)


    . . . . . . . . . . . . . . . . . . . . .


    (x^5 + x^3 + x) + (x^4 + x^2 + 1) =

    x(x^4 + x^2 + 1) + 1(x^4 + x^2 + 1) =

    (x^4 + x^2 + 1)(x + 1) =

    (x^2 - x + 1)(x^2 + x + 1)(x + 1)

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